2.) Solve for \( \left(0^{\circ}<\theta<90^{\circ}\right) \) 1) \( \sin ^{2} \theta-\frac{1}{2} \sin \theta=0 \)

To tackle the equation \( \sin^{2} \theta - \frac{1}{2} \sin \theta = 0 \), you can factor it: \( \sin \theta(\sin \theta - \frac{1}{2}) = 0 \). This yields two potential solutions: \( \sin \theta = 0 \) or \( \sin \theta = \frac{1}{2} \). Since we're looking for \( \theta \) in the range \( 0^{\circ} < \theta < 90^{\circ} \), the solution \( \sin \theta = 0 \) isn't applicable. Thus, \( \sin \theta = \frac{1}{2} \) gives \( \theta = 30^{\circ} \). Now, if you ever wondered why sine values are important, they're not just for angles—they're key in real-world applications like engineering and physics. For instance, they help you understand wave patterns or the forces acting on structures! Another handy tip: be sure to check that you can't overlook solutions in other quadrants! While \( \sin \theta = \frac{1}{2} \) gives you the familiar \( 30^{\circ} \) in the first quadrant, don't forget about \( 150^{\circ} \) in the second quadrant when venturing outside our given range. Always stay aware of the angles your trig functions can create!